ODE Homework 12.1. Linear Equations; Method of Integrating Factors

1. Find the solution of the given initial value problem

y′ +2

ty =

cos t

t2, y(π) = 0, t > 0

[§2.1 #16]Sol. The integrating factor µ(t) is

µ(t) = exp

∫2

tdt = e2 ln |t| = t2

Multiple the equation by µ(t), we get

t2y′ + 2ty = (t2y)′ = cos t

Integrating both side of the equation, we have that

t2y = sin t+ C ⇒ y(t) =sin t

t2+ Ct−2

for some constant C. Since y(π) = 0, so

0 =sin π

π2+C

π2⇒ C = 0

Hence the specific solution of the initial value problem is

y(t) =sin t

t2

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2. Find the solution of the given initial value problem

ty′ + (t+ 1)y = t, y(ln 2) = 1, t > 0

[§2.1 #20]Sol. Rewrite the equation as the form

y′ +t+ 1

ty = 1

The integrating factor µ(t) is

µ(t) = exp

∫t+ 1

tdt = et+ln |t| = tet

since t > 0. Multiple the equation by µ(t), we get

tety′ + (t+ 1)ety = (tety)′ = tet

Integrating both side of the equation, we have that

tety =

∫tetdt = (t− 1)et + C ⇒ y(t) =

t− 1

t+Ce−t

t

for some constant C. Since y(ln 2) = 1, so

1 =ln 2− 1

ln 2+Ce− ln 2

ln 2⇒ C = 2

Hence the specific solution of the initial value problem is

y(t) =t+ 2e−t − 1

t

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3. Consider the initial value problem

y′ +1

4y = 3 + 2 cos 2t, y(0) = 0.

(a) Find the solution of this initial value problem and describeits behavior for large t.

(b) Determine the value of t for which the solution first inter-sects the line y = 12.

[§2.1 #29]Sol.(a) The integrating factor µ(t) = exp

( ∫14dt)

= et4 . And the

equation can be written as

et4y′ +

1

4e

t4y = (e

t4y)′ = 3e

t4 + 2e

t4 cos 2t

Integrating both side, we can conclude that the generalsolution for the equation is

y(t) = Ce−t4 +

8

65cos 2t+

64

65sin 2t+ 12

Together with the initial condition y(0) = 0, we get thesolution for the initial value problem is

y(t) = −788

65e−

t4 +

8

65cos 2t+

64

65sin 2t+ 12

= −788

65e−

t4 +

8√65

sin(2t+ θ) + 12

where θ = sin−1 6465

. As t→∞, we have that

y(t)→ 8√65

sin(2t+ θ) + 12

which means the solution will oscillate about an average12,with an amplitude of 8√

65.

(b) By letting y(t) = 12, then we have the equality

8 cos 2t+ 64 sin 2t− 788e−t4 = 0

Let g(t) = 8 cos 2t + 64 sin 2t − 788e−t4 , then g(10) ≈

−2.99 < 0, g(10.3) ≈ 1.53 > 0. By using Newton’s methodwith initial point x0 = 10, we have that the first value t forwhich y(t) intersects y = 12 is t ≈ 10.066.

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4. Consider the initial value problem

y′ − 3

2y = 3t+ 2et, y(0) = y0

Find the value of y0 that separates solutions that grow positivelyas t → ∞ from those that grow negatively. How does thesolution that corresponds to this critical value of y0 behave ast→∞?[§2.1 #31]

Sol. The integrating factor µ(t) = exp( ∫−3

2dt)

= e−3t2 . And

the equation can be written as

e−3t2 y′ − 3

2e

−3t2 y = (e

−3t2 y)′ = 3te−

3t2 + 2e−

t2

Integrating both side, we have that the general solution for theequation is

y(t) = Ce3t2 − 4et − 2t− 4

3

The initial condition y(0) = y0 implies that the specific solutionfor the initial value problem is

y(t) =(y0 +

16

3

)e

3t2 − 4et − 2t− 4

3

As t→∞, the term(y0 + 16

3

)e

3t2 will dominate the solution. Its

sign will determine the divergence properties. Hence the criticalvalue of the initial condition is y0 = −16

3. The corresponding

solution y(t) = −4et−2t− 43, will also decrease without bound.

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5. Show that all solutions of 2y′ + ty = 2 approach a limit ast→∞, and find the limiting value.[§2.1 #32]

Proof. Rewrite the equation as y′ + t2y = 1. The integrating

factor µ(t) = exp( ∫

t2dt)

= et2

4 . And the equation can bewritten as

et2

4 y′ +t

2e

t2

4 y =(e

t2

4y)′

= et2

4

Integrating both side, we have that the general solution for theequation is

y(t) =

∫ t

es2−t2

4 ds

Thus, by L’Hospital Rule,

limt→∞

y(t) = limt→∞

∫ t

es2−t2

4 ds = limt→∞

∫ te

s2

4 ds

et2

4

= limt→∞

et2

4

t2e

t2

4

= limt→∞

2

t= 0

Hence we are done. �

6. Variation of Parameters. Consider the following method ofsolving the general linear equation of first order:

y′ + p(t)y = g(t). (i)

(a) If g(t) = 0 for all t, show that the solution is

y = A exp[−∫p(t)dt

], (ii)

where A is a constant.(b) If g(t) is not everywhere zero, assume that the solution of

Eq. (i) is of the form

y = A(t) exp[−∫p(t)dt

], (iii)

where A is now a function of t. By substituting for y inthe given differential equation, show that A(t) must satisfythe condition

A′(t) = g(t) exp[ ∫

p(t)dt]. (iv)

(c) Find A(t) from Eq. (iv). Then substitute for A(t) in Eq.(iii) and determine y. Verify that the solution obtainedin this manner agrees with that of Eq. (33) in the text.This technique is known as the method of variation ofparameters.

[§2.1 #38]

Proof.(a) If g(t) = 0 for all t, then the equation becomes y′+p(t)y =

0. The integrating factor is µ(t) = exp[ ∫ t

p(s)ds]. And

the equation can be written as

exp[ ∫ t

p(s)ds]y′+p(t) exp

[ ∫ t

p(s)ds]y =

(exp

[ ∫ t

p(s)ds]y)′

= 0

Integrating both side, we have that the general solution forthe equation is

y(t) = A exp[−∫ t

p(s)ds]

for some constant A.(b) For the case g(t) 6≡ 0, assume that the solution of Eq.

(i) is of the form y(t) = A(t) exp[−∫ tp(s)ds

]. Since

y′ + p(t)y = g(t), so

A′(t) exp[−∫ t

p(s)ds]

= g(t)

Hence

A′(t) = g(t) exp[ ∫ t

p(s)ds]

(c) Since A′(t) = g(t) exp[ ∫ t

p(s)ds], we have that

A(t) =

∫ t

g(s) exp[ ∫ s

p(τ)dτ]ds =

∫ t

g(s)e∫ s p(τ)dτds

Hence

y(t) = A(t) exp[−∫ t

p(s)ds]

=

∫ t

g(s)e∫ s p(τ)dτds · e−

∫ t p(s)ds

=

∫ t

g(s)e∫ st p(τ)dτds

By letting µ(t) = exp[ ∫ t

p(s)ds], then the expression of

y(t) is exactly the same with Eq. (33) in the text book.�

7. Use the method of variation of parameters to solve the givendifferential equation.

ty′ + 2y = cos t, t > 0

[§2.1 #41]Sol. Rewrite the equation as the form y′ + 2

ty = cos t

t. The

integrating factor µ(t) = exp( ∫

2tdt)

= t2. Hence the general

solution of the homogeneous equation y′ + 2ty = 0 is

y0(t) = A exp[−∫

2

tdt]

=A

t2

for some constant A. Assume that the solution for the original

equation is of the form y(t) = A(t)t2

. Substituting y(t) into theequation we get

A′(t) =cos t

tµ(t) = t cos t

and thus A(t) =∫t cos tdt = t sin t+cos t+C for some constant

C. Therefore, the solution y(t) of the equation is

y(t) =A(t)

t2=t sin t+ cos t+ C

t2

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2.2. Separable Equations

8. Solve the given differential equation.

dy

dx=x− e−x

y + ey

[§2.2 #7]Sol. Rewrite the equation of the form (y+ey)dy = (x−e−x)dx.Then∫

(y + ey)dy =

∫(x− e−x)dx⇒ ey +

y2

2= e−x +

x2

2+ C

for some constant C, which defines the solution y(x) of theequation implicitly. �

9. Solve the initial value problem

y′ =3x2

3y2 − 4, y(1) = 0

and determine the interval in which the solution is valid.[§2.2 #22]Sol. Rewrite the equation of the form (3y2 − 4)dy = 3x2dx.Then∫

(3y2 − 4)dy = 3

∫x2dx⇒ y3 − 4y = x3 + C

for some constant C, which defines the general solution y(x)of the equation implicitly. Since y(1) = 0, so we have that1 + C = 0 ⇒ C = −1. Hence the solution y(x) for the initialvalue problem is defined implicitly by the equation

y3 − 4y = x3 − 1

Note that the differential equation y′ = 3x2

3y2−4 implies that the

slope of tangent lines tends to infinity as y → ± 2√3. Substitut-

ing y = ± 2√3

into y3 − 4y = x3 − 1 we get x3 − 1 = ∓ 163√3. By

solving above two equations, we have that x ≈ −1.276 and 1.598respectively. Therefore, the solution y(x) is valid for about theinterval (−1.276, 1.598). �

10. Solve the initial value problem

y′ =2 cos 2x

3 + 2y, y(0) = −1

and determine where the solution attains its maximum value.[§2.2 #25]Sol. Rewrite the equation of the form (3 + 2y)dy = 2 cos 2xdx.Then∫

(3 + 2y)dy = 2

∫cos 2xdx⇒ y2 + 3y = sin 2x+ C

for some constant C, which defines the general solution y(x) ofthe equation implicitly. Since y(0) = −1, so have that C = −2so that y2+3y = sin 2x−2. Furthermore, the quadratic formulatogether with the initial condition y(0) = −1 implies that

y(x) = −3

2+

√sin 2x+

1

4

Note that y′(x) = cos 2x√sin 2x+ 1

4

, we have that the solution is valid

for sin 2x > −14, together with the initial condition y(0) = −1,

the solution is valid on interval I =(

sin−1(−14), π

2−sin−1(−1

4)).

Thus x = π4

is the only critical point in the interval I. Observethat

y′′(π

4

)= − 4√

5< 0

Then y(x) has maximum at x = π4, which is y

(π4

)=√5−32

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11. Consider the initial value problem

y′ =ty(5− y)

1 + t, y(0) = y0 > 0.

(a) Determine how the solution behaves as t→∞.(b) If y0 = 2, find the time T at which the solution first reaches

the value 4.99.(c) Find the range of initial values for which the solution lies

in the interval 4.99 < y < 5.01 by the time t = 2.[§2.2 #28]Sol.(a) Rewrite the equation of the form dy

y(5−y) = tdt1+t

. Then∫dy

y(5− y)=

∫tdt

1 + t⇒ ln |y| − ln |5− y| = 5t− 5 ln |1 + t|+ C0

⇒ ln∣∣∣ y

5− y

∣∣∣ = 5t− 5 ln |1 + t|+ C0

⇒∣∣∣ y

5− y

∣∣∣ =Ce5t

(1 + t)5

where C = |y0||5−y0| according to the initial condition y(0) =

y0 > 0. As t→∞, we have that∣∣∣ y

5− y

∣∣∣→ limt→∞

Ce5t

(1 + t)5= lim

t→∞

55Ce5t

5!=∞

Hence∣∣ y5−y

∣∣ =∣∣ 55−y − 1

∣∣→∞ which implies that y(t)→ 5as t→∞.

(b) If y(0) = 2, then C = 25−2 = 2

3, and hence the solution for

the initial value problem is defined implicitly by

y

5− y=

2e5t

3(1 + t)5

Since ddt

(2e5t

3(1+t)5

)= 10te5t

3(1+t)6≥ 0 for all t ≥ 0, we have that

y5−y > 0 for all t ≥ 0. According to the differential equation

itself, we know that y′ is always positive and hence y(t) isincreasing for all t ≥ 0. To obtain T , we simply solve theequation

4.99

5− 4.99=

2e5T

3(1 + T )5

to get T ≈ 2.6063.

(c) Note that y(t) = 5 is an equilibrium solution. Accordingto (b), we know that if y0 < 5, then y(t) < 5, ∀ t ≥ 0 andsimilarly, if y0 > 5, then y(t) > 5, ∀ t ≥ 0. Hence that thegeneral solution for the equation is defined implicitly by

y

5− y=

y0e5t

(5− y0)(1 + t)5, y0 6= 5

For t = 2, if 4.99 < y < 5.01, then y5−y < −499 or y

5−y >

501. Thus we have that

y05− y0

< −499 · 35

e10or

y05− y0

>501 · 35

e10

This implies that the initial value y0 lies about approxi-mately in the interval (4.23397, 6.10986).

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12. Consider the equation

dy

dx=y − 4x

x− y. (i)

(a) Show that Eq. (i) can be rewritten as

dy

dx=

(y/x)− 4

1− (y/x); (ii)

thus Eq. (i) is homogeneous.(b) Introduce a new dependent variable v so that v = y/x, or

y = xv(x). Express dydx

in terms of x, v and dvdx

.

(c) Replace y and dydx

in Eq. (ii) by the expressions from part

(b) that involve v and dvdx

. Show that the resulting differ-ential equation is

v + xdv

dx=v − 4

1− v,

or

xdv

dx=v2 − 4

1− v, (iii)

Observe that Eq. (iii) is separable.(d) Solve Eq. (iii), obtaining v implicitly in terms of x.(e) Find the solution of Eq. (i) by replacing v by y/x in the

solution in part (d).(f) Draw a direction field and some integral curves of Eq.(i).

Recall that the right side of Eq. (i) actually depends onlyon the ratio y/x. This means that integral curves have thesame slope at all points on any given straight line through

the origin, although the slope changes from one line to an-other. Therefore the direction field and the integral curvesare symmetric with respect to the origin. Is this symmetryproperty evident from your plot?

[§2.2 #30]Sol.(a) By dividing x on both the numerator and denominator of

the right hand side of Eq. (i), we get Eq. (ii).(b) Since y = vx, so dy

dx= v + x dv

dx.

(c) Since dydx

= (y/x)−41−(y/x) = v−4

1−v , by (b), we get

v + xdv

dx=v − 4

1− v⇒ x

dv

dx=v2 − 4

1− v(d) Note that Eq. (iii) is an separable equation, we can rewrite

the equation as the form 1−vv2−4dv = dx

x. Then∫

1− vv2 − 4

dv =

∫dx

x⇒ ln |2− v|+ 3 ln |2 + v| = −4 ln |x|+ C0

⇒ ln |(2− v)(2 + v)3| = ln1

x4+ C0

⇒ x4|v − 2||v + 2|3 = C

where C = eC0 for some constant C0. Thus the generalsolution v(x) for Eq. (iii) is defined implicitly by the equa-tion

x4|v − 2||v + 2|3 = C

(e) Note that y = vx, so the general solution y(x) of Eq. (i) isdefined implicitly by

(y − 2x)(y + 2x)3 = C ⇒ y4 + 4xy3 − 16x3y − 16x4 = C

(f) The plot of the direction field and integral curves of Eq.(i) is as follows

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13. Given differential equation

(x2 + 3xy + y2)dx− x2dy = 0

(a) Show that the given equation is homogeneous.(b) Solve the differential equation.(c) Draw a direction field and some integral curves. Are they

symmetric with respect to the origin?[§2.2 #36]Sol.(a) Rewrite the equation as the form

dy

dx=x2 + 3xy + y2

x2= 1 + 3

(yx

)+(yx

)2which implies that the equation is homogeneous.

(b) Let v = yx, then the above equation becomes

v + xdv

dx= 1 + 3v + v2 ⇒ x

dv

dx= (v + 1)2

Thus, dv(v+1)2

= dxx

, and hence∫dv

(v + 1)2=

∫dx

x⇒ −1

v + 1= ln |x|+ C

⇒ v =1

C − ln |x|− 1

Replacing v = yx, we got the general solution y(x) for the

original equation that y = xC−ln |x| − x.

(c) The plot of the direction field and integral curves for theequation is as follows Clearly, the integral curves are sym-

metric with respect to the origin.�

2.5. Autonomous Equations and Population Dynamics

14. Consider the equation of the form dydt

= f(y), y0 ≥ 0, wheref(y) = y(y − 1)(y − 2). Sketch the graph of f(y) versus y.Determine the critical (equilibrium) points, and classify it asasymptotically stable or unstable. Draw the phase line, andsketch several graphs of solutions in the ty-plane.[§2.5 #3]Sol. The graph of f(y) is as follows. Since the critical points

occurs at dydt

= 0, it is easy to see that y = 0, 1, 2 are criticalpoints. That is, y = 0, 1, 2 are equilibrium solutions for theequation. Note that f(y) is positive on (0, 1) ∪ (2,∞), while itis negative on (−∞, 0)∪ (1, 2). We see that the solution y(t) isincreasing for 0 < y < 1 and y > 0, while decreasing for y < 0and 1 < y < 2. Hence the critical points y = 0 and y = 2 areunstable, while the critical point y = 1 is asymptotically stable.The graph of solutions and the phase line is as follows

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15. Consider the equation of the form dydt

= f(y), −∞ < y0 < ∞,where f(y) = y2(y2 − 1). Sketch the graph of f(y) versus y.Determine the critical (equilibrium) points, and classify it asasymptotically stable or unstable. Draw the phase line, andsketch several graphs of solutions in the ty-plane.[§2.5 #9]Sol. The graph of f(y) is as follows

Since the critical points occurs at dydt

= 0, it is easy to see thaty = 0, ±1 are critical points. That is, y = 0, ±1 are equilib-rium solutions for the equation. Note that f(y) is positive on(−∞,−1) ∪ (1,∞), while it is negative on (−1, 0) ∪ (0, 1). Wesee that the solution y(t) is increasing for y < −1 and y > 1,while decreasing for −1 < y < 0 and 0 < y < 1. Hence thecritical points y = 0 is semistable, while the critical point y = 1is unstable, and the critical point y = −1 is asymptotically sta-ble. The graph of solutions and the phase line is as follows:

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16. Suppose that a certain population obeys the logistic equationdydt

= ry[1− y

K

].

(a) If y0 = K4

, find the time τ at which the initial populationhas doubled. Find the value of τ corresponding to r =0.025 per year.

(b) If y0K

= α, find the time T at which y(T )K

= β, where 0 <α, β < 1. Observe that T → ∞ as α → 0 or as β → 1.Find the value of T for r = 0.025 per year, α = 0.1, andβ = 0.9.

[§2.5 #15]Sol.(a) Observe that the logistic equation is separable. By solving

the equation, we get the general solution is

y(t) =y0K

y0 + (K − y0)e−rt

where y0 = y(0) is the initial condition. We can write t asa function of y,

t =−1

rln∣∣∣y0(K − y)

y(K − y0)

∣∣∣With y0 = K

4, we have that t = −1

rln∣∣K−y

3y

∣∣. Setting y =

2y0 = K2

, we get

τ =−1

rln∣∣∣K − K

23K2

∣∣∣ =ln 3

r

If r = 0.025 per year. τ ≈ 43.9445 year.

(b) By letting α = y0K

and β = y(T )K

= yK

, then

T =−1

rln∣∣∣α(1− β)

β(1− α)

∣∣∣If r = 0.0025 per year and α = 0.1, β = 0.9, then T ≈175.778 year.

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17. Another equation that has been used to model population growthis the Gompertz equation

dy

dt= ry ln

(Ky

),

where r and K are positive constants.

(a) Sketch the graph of f(y) versus y, find the critical points,and determine whether each is asymptotically stable orunstable.

(b) For 0 ≤ y ≤ K, determine where the graph of y versus t isconcave up and where it is concave down.

(c) For each 0 < y ≤ K, show that dydt

as given by the Gom-

pertz equation is never less than dydt

as given by the logisticequation.

[§2.5 #16]Sol.(a) Let f(y) = ry ln

(Ky

), then f ′(y) = r

(ln K

y− 1

). It is

clear that f(0) = f(K) = 0 and f ′(y) = 0 if and only ify = K

e< K. The graph of f(y) versus y is as follows

The critical points are y = 0 and y = K. Since f(y) ispositive on (0, K) and negative on (K,∞), the solution y(t)is increasing for 0 < y < K and is decreasing for y > K.Thus y = 0 is unstable and y = K is asymptotically stable.

(b) Note that for 0 ≤ y ≤ K, dydt

= f(y) > 0. By chain rule,

d2y

dt2=

d

dt

(dydt

)=

d

dtf(y) = f ′(y)

dy

dt

Observe that if 0 ≤ y ≤ Ke, f(y) is increasing while if

Ke≤ y ≤ K, f(y) is decreasing. Thus f ′(y) > 0 for 0 ≤

y ≤ Ke

and f ′(y) < 0 for Ke≤ y ≤ K. This implies that

the solution y(t) is concave up for 0 ≤ y ≤ Ke

while it is

concave down for Ke≤ y ≤ K.

(c) Let g(y) = ry(1 − y

K

)and set h(y) = f(y)−g(y)

ry. Then

h(K) = 0, and

h′(y) =yf ′(y)− yg′(y)− f(y) + g(y)

ry2=

1

K− 1

y≤ 0

since 0 < y ≤ K. This shows that f(y) ≥ g(y) for 0 <y ≤ K. Therefore, dy

dt= ry ln

(Ky

)is never less than dy

dt=

ry(1− y

K

)for 0 < y ≤ K.

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18. Harvesting a Renewable Resource. Suppose that the papu-lation y of a certain species of fish (for example, tuna or halibut)in a given area of the ocean is described by the logistic equation

dy

dt= r(

1− y

K

)y.

Although it is desirable to utilize this source of food, it is in-tuitively clear that if too many fish are caught, then the fishpopulation may be reduced below a useful level and possiblyeven driven to extinction. At a given level of effort, it is reason-able to assume that the rate at which fish are caught dependson the population y: the more fish there are, the easier it is tocatch them. Thus we assume that the rate at which fish arecaught is given by Ey, where E is a positive constant, withunits of 1/times, that measures the total effort made to har-vest the given species of fish. To include this effect, the logisticequation is replaced by

dy

dt= r(

1− y

K

)y − Ey.

This equation is known as the Schaefer model after the biol-ogist M. B. Schaefer, who applied it to fish populations.(a) Show that if E < r, then there are two equilibrium points

y1 = 0 and y2 = K(1− E

r

)> 0.

(b) Show that y = y1 is unstable and y = y2 is asymptoticallystable.

(c) A sustainable yield Y of the fishery is a rate at which fishcan be caught indefinitely. It is the product of the effortE and the asymptotically stable population y2. Find Yas a function of the effort E; the graph of this function isknown as the yield-effort curve.

(d) Determine E so as to maximize Y and thereby find themaximum sustainable yield Ym.

[§2.5 #20]Sol.(a) Rewrite the equation of the form

dy

dt= y(r − E − ry

K

)Then dy

dt= 0 if and only if y = 0 or

r − E − ry

K= 0⇒ y = K

(1− E

r

)Hence if E < r, then there are two equilibrium pointsy1 = 0 and y2 = K

(1− E

r

)> 0.

(b) Let f(y) = y(r − E − ry

K

), then f(y) > 0 for 0 < y < y2

and f(y) < 0 for y > y2. This implies that the solutiony(t) is increasing for 0 < y < y2 and is decreasing fory > y2. Therefore y = y1 = 0 is unstable and y = y2 isasymptotically stable.

(c) It is easy to see that

Y = y2E = KE(

1− E

r

)(d) Note that

Y = KE − KE2

r= −K

r

(E − r

2

)2+rK

4Hence when E = r

2, we have maximum sustainable yield

Ym =rK

4�

19. Bifurcation Points. For an equation of the form

dy

dt= f(a, y), (i)

where a is a real parameter, the critical points (equilibriumsolutions) usually depend on the value of a. As a steadily in-creases or decreases, it often happens that at a certain value ofa, called a bifurcation point, critical points come together,or separate, and equilibrium solutions may either be lost orgained. Bifurcation points are a great interest in many applica-tions, because near them the nature of the solution of the un-derlying differential equation is undergoing an abrupt change.For example, in fluid mechanics a smooth (laminar) flow maybreak up an become turbulent. Or an axially loaded column

may suddenly buckle and exhibit a large lateral displacement.Or, as the amount of one of the chemicals in a certain mixtureis increased, spiral wave patterns of varying color may suddenlyemerge in an originally quiescent fluid. The following exerciseconsider the equation

dy

dt= a− y2. (ii)

(a) Find all of the critical points for Eq. (ii). Observe thatthere are no critical points if a < 0, one critical point ifa = 0, and two critical points if a > 0.

(b) Draw the phase line in each case and determine whethereach critical point is asymptotically stable, semistable, orunstable.

(c) In each case sketch several solutions of Eq. (ii) in the ty-plane.

(d) If we plot the location of the critical points as a function ofa in the ay-plane, we obtain Figure 2.5.10. This is calledthe bifurcation diagram for Eq. (ii). The bifurcation ata = 0 is called a saddle-node bifurcation. This name ismore natural in the context of second order systems.

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2.5 Autonomous Equations and Population Dynamics 93

fluid. Problems 25 through 27 describe three types of bifurcations that can occur in simpleequations of the form (i).25. Consider the equation

dy/dt = a − y2. (ii)

(a) Find all of the critical points for Eq. (ii). Observe that there are no critical points ifa < 0, one critical point if a = 0, and two critical points if a > 0.(b) Draw the phase line in each case and determine whether each critical point is asymp-totically stable, semistable, or unstable.(c) In each case sketch several solutions of Eq. (ii) in the ty-plane.(d) If we plot the location of the critical points as a function of a in the ay-plane, we obtainFigure 2.5.10. This is called the bifurcation diagram for Eq. (ii). The bifurcation at a = 0is called a saddle–node bifurcation. This name is more natural in the context of secondorder systems, which are discussed in Chapter 9.

–2

–1

1

2

–2 –1 1 2 3 4

Unstable

Asymptotically stable

y

a

FIGURE 2.5.10 Bifurcation diagram for y′ = a − y2.

26. Consider the equationdy/dt = ay − y3 = y(a − y2). (iii)

(a) Again consider the cases a < 0, a = 0, and a > 0. In each case find the critical points,draw the phase line, and determine whether each critical point is asymptotically stable,semistable, or unstable.(b) In each case sketch several solutions of Eq. (iii) in the ty-plane.(c) Draw the bifurcation diagram for Eq. (iii), that is, plot the location of the critical pointsversus a. For Eq. (iii) the bifurcation point at a = 0 is called a pitchfork bifurcation; yourdiagram may suggest why this name is appropriate.

27. Consider the equationdy/dt = ay − y2 = y(a − y). (iv)

(a) Again consider the cases a < 0, a = 0, and a > 0. In each case find the critical points,draw the phase line, and determine whether each critical point is asymptotically stable,semistable, or unstable.(b) In each case sketch several solutions of Eq. (iv) in the ty-plane.

[§2.5 #25]Sol.(a) Let dy

dx= 0, we have y2 = a. Then there are no critical

points if a < 0, there is one critical point y = 0 if a = 0,and there are two critical points y = ±

√a if a > 0.

(b) For a < 0, there is no critical point. For a = 0, the criticalpoint y = 0 is semistable. For a > 0 the critical pointy =√a is asymptotically stable and y = −

√a is unstable.

The phase lines of each cases is as follows

a < 0 a = 0 a > 0

(c) The graph of solutions for each case is as follows

a < 0 a = 0 a > 0

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